Question

1. Give a condition that is necessary but not sufficient for an undirected graph to have an Eulerian Path. Justify your answer.

2. Give a condition that is sufficient but not necessary for an undirected graph not to have an Eulerian Cycle. Justify your answer.

Answer 1

Eulerian Path An undirected graph has Eulerian Path if following two conditions are true. ...a) Same as condition (a) for Eulerian Cycle ...b) If two vertices have odd degree and all other vertices have even degree. Note that only one vertex with odd degree is not possible in an undirected graph (sum of all degrees is always even in an undirected graph) Note that a graph with no edges is considered Eulerian because there edges to traverse.

Eulerian Cycle An undirected graph has Eulerian cycle if following two conditions are true. ....a) All vertices with non-zero degree are connected. We don't care about vertices with zero degree because they don't belong to Eulerian Cycle or Path (we only consider all edges) ...b) All vertices have even degree.

1. Give the necessary and sufficient conditions for an undirected connected graph to have an Eulerian walk Solution (a) Prove: If an undirected graph G has an Eulerian walk W, the graph can have at most two odd degree vertices. For a vertex v, let d(v) be the degree of v and let nw (v) be the number of edges on W incident to v. Since W is Eulerian, n(v) = d(v). Also observe that nw(v) must be even for all vertices other than the start and end vertices of the W. This is because each time W enters an intermediate vertex, it must exit it so W uses up two edges incident to it. Therefore, there can be at most two odd degree vertices (the start and end vertex) (b) Prove: If a connected graph has at most two odd degree vertices, it has an Eulerian walk We learned in lecture that there must be an even number of odd degree vertices, since the sum of the degrees of all vertices must be even. Therefore, there are either 0 odd degree vertices or 2 odd degree vertices. If there are 0, we have an Eulerian tour by Euler's theorem. If there are 2 odd degree vertices, u and v, we can construct an Eulerian walk We will start our walk T from vertex u, never repeating edges. Let's say we get stuck at vertex w. If w u, then nT(w) is even, since we started at u. But since d(u) is odd, there must exist at least one unused edge incident to u which we can use to continue. If w # u, then nT(w) is odd we have entered w but not exited. If w is even degree, there must exist at least one unused edge incident to w which we can use to exit. Therefore, w must be odd degree, so w must be v We've created a walk T, but it is not necessarily Eulerian. If we remove T from the graph the remaining graph has even degree. This is because we are subtracting nT (w) from the degree of each vertex w. For the even degree vertices (vertices other than u and v), we are modifying the degree by subtracting an even number, resulting in an even degree. For u and v we are subtract- ing odd numbers from their degrees, resulting in even degrees. Now we can construct Eulerian tours in the remaining graph. These tours might be disconnected, but they can all be connected to T. This is due to Claim 3 from note 7 (we can generalize Claim 3 so that A is a trail and not a tour- the proof remains the same). Finally, we splice together T with all the Eulerian tours to create the final Eulerian trail